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I feel silly asking such elementary questions, but hopefully this is appropriate for math.stackexchange.

I'm studying to take calculus next semester but I haven't done any math in a long time, so I've been trying to brush up on my algebra and my trig. I'm stuck on this problem here:

$$ (4 x-2)^2-2 (4 x-2) = 15 $$

I was able to get to this step:

$$ 8-24 x+16 x^2 = 15 $$

But after I'm confused on what to do after this step.

Also(a related question): I got some help with this equation earlier:

$$(u+1)/(u+4)+1 = (u-5)/(u-4)$$ The last step ended up being $$x (x -2) = 0$$ The explanation I recieved was that "this implied that $x = 0$ or $x = 2$" I was wondering if someone can help me understand why this is so, I feel as if I'm missing a vital piece of information preventing me from understanding these problems.

I'd be much obliged if someone can point me in the right direction.

Here are the wolfram links for both equations:

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It's called the zero product property, and is a result of the real numbers forming an integral domain. – The Chaz 2.0 Nov 1 '11 at 1:30
Thanks for the vocab lesson, a quick googling turned up some useful links. – lachance Nov 1 '11 at 1:36
No problem. There's some (relatively) advanced stuff in ring theory, but for now the answers that you have should suffice. – The Chaz 2.0 Nov 1 '11 at 1:40
up vote 5 down vote accepted

It's one of the basic features of numbers that if you multiply two numbers and the result is 0, then at least one of the factors must have been 0. (Namely if $ab=0$ and $a$ is not $0$, then we're allowed to divide both sides of the equation by $a$ and get $b=\frac0a=0$. So either $a$ is $0$ or $b$ is).

Therefore $x(x-2)=0$ implies that $x=0$ or $x-2=0$ and the latter is just a different way to write $x=2$.

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Thanks for this, I've always been taught to "just take this and that set it to 0" with no good explanation why. This cleared up some gaps in my understanding. Any chance I could get some help on that first equation? I'm really stuck. – lachance Nov 1 '11 at 1:43
Oh, there was a first question. Sorry, I read too fast. Subtract 15 from both sides to get $16x^2-24x-7=0$. Then use the quadratic formula. – Henning Makholm Nov 1 '11 at 1:48
O_O... Oh man, I was looking for it to work out like the second equation so much, flipping it every which way, I totally forgot about the quadratic equation. Thanks again! >_< – lachance Nov 1 '11 at 1:54
Once you do find the roots ($-\frac14$ and $\frac74$) you can check by multiplying through that $16x^2-24x-7=16(x+\frac14)(x-\frac74)$, in which case the setting-zero trick actually does work. But the easiest way to find this factorization is ... to use the quadratic formula to find the roots! – Henning Makholm Nov 1 '11 at 2:06
Thank ou for your patience and your time. You are a true gentleman and a scholar :) – lachance Nov 1 '11 at 2:16

We consider only the problem of finding the solutions of $(4 x-2)^2-2 (4 x-2) = 15$. You multiplied out correctly. We will look at the equation you got a little later. But first, we do the problem in a more lazy (and efficient) way.

It seems natural to let $u=4x-2$. Then our equation can be rewritten as $$u^2-2u-15=0.$$ Possibly we can find the solutions by inspection. Or else we can factor the quadratic, obtaining $$u^2-2u-15=(u-5)(u+3).$$ Note that $(u-5)(u+3)=0$ when $u=5$, when $u=-3$, and for no other value of $u$. It follows that $x$ is a solution of the original equation when $4x-2=-3$ and when $4x-2=5$, and for no other value of $x$.

We now have a couple of linear equations for the roots. To solve $4x-2=-3$, rewrite as $4x=(-3)+2=-1$, and divide both sides by $4$ to get $x=-\dfrac{1}{4}$. Similarly, the equation $4x-2=5$ has the solution $x=\dfrac{7}{4}$.

Instead of factoring $u^2-2u-15$, we could use the Quadratic Formula. For details, please see below.

Another way: Expand, as you did, and bring all terms to the left-hand side. We obtain the equation $$16 x^2-24x-7 =0.$$ It turns out that the quadratic polynomial on the left can be factored without too much trouble. (Many quadratic polynomials in school exercises can be factored in a simple way. One cannot count on that in the real world!) So we recall the standard formula for solving the quadratic equation $ax^2+bx+c=0$ (where $a \ne 0$). The solutions are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ For a derivation of this formula, any source will do. I even have one on Stack Exchange. The formula is so frequently useful that most students remember it. Now calculate, with $a=16$, $b=-24$, and $c=-7$. It turns out that $b^2-4ac=1024=(32)^2$, so the roots are not complicated.

Comment: Let $P(x)$ be a polynomial with integer coefficients. Suppose that the highest degree term is $ax^n$, and the constant term is $c$. Then any rational root of $P(x)=0$ must be of the form $p/q$, where $p$ is a (possibly negative) factor of $c$, and $q$ is a positive factor of $a$. (This result is often called the Rational Roots Theorem.)

So if we are looking for the rational roots of $16x^2-24x-7=0$, the only possible candidates are $\pm\frac{1}{1}$, $\pm\frac{1}{2}$, $\pm\frac{1}{4}$, $\pm\frac{1}{8}$, $\pm\frac{1}{16}$, $\pm\frac{7}{1}$, $\pm\frac{7}{2}$, $\pm\frac{7}{4}$, $\pm\frac{7}{8}$, and $\pm\frac{7}{16}$. We can just try all these, to see which, if any, work. It isn't quite as unpleasant as it looks. However, it is not really a suitable approach to quadratic equations. After all, the roots might not be rational. And the Quadratic Formula is simple to use.

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A full, very concise and understandable explanation. I need to practice a few more problems just to make sure but the concept seems clear. There are no words to express my gratitude. Thank you. – lachance Nov 1 '11 at 3:21

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